A rocket is launched straight up with constant acceleration. Four seconds after liftoff, a bolt falls off the side of the rocket. The bolt hits the ground 6.70 s later. Find the rockets acceleration. I dont understand how to set up this problem, ive never done something like this , asked some friends and they didnt know how to do it. Any help would be greatly appreciated
Work backwards kind of. If it took 6.7 for the bolt to hit the ground, how high was it? [tex]d = v_{i}t + \frac{1}{2}at^2[/tex] vi = 0 a = -32 ft\sec^2 or -9.8 m\sec^2 t = 6.7 s Solve for d. Now, use the same equation and solve for "a" of the rocket. [tex]d = v_{i}t + \frac{1}{2}at^2[/tex] d = part a vi = 0 t = 4 Jameson
Do it in two parts, but like this: Part 1: Find the height and speed of the rocket after 4 seconds in terms of the rocket acceleration. Use: [itex]h = 1/2 a t^2[/itex] and [itex]v = a t[/itex] Part 2: Realize that the initial height and speed of the bolt equals the height and speed of the rocket after 4 seconds (which was calculated in part 1). The height of the bolt after it leaves the rocket is: [itex]h = h_0 + v_0t -(g/2)t^2[/itex], using [itex]h_0[/itex] and [itex]v_0[/itex] (in terms of a) from part 1; set h = 0 when t = 6.7 seconds. Solve for a.